YES
The TRS could be proven terminating. The proof took 1285 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (46ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (306ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| | – Problem 6 was processed with processor DependencyGraph (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor PolynomialLinearRange4iUR (853ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| if_minus#(false, s(x), y) | → | minus#(x, y) | | quot#(s(x), s(y)) | → | minus#(x, y) |
| log#(s(s(x))) | → | quot#(x, s(s(0))) | | le#(s(x), s(y)) | → | le#(x, y) |
| minus#(s(x), y) | → | le#(s(x), y) | | quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
| minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) | | log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, true, false, quot, log
Strategy
The following SCCs where found
| if_minus#(false, s(x), y) → minus#(x, y) | minus#(s(x), y) → if_minus#(le(s(x), y), s(x), y) |
| le#(s(x), s(y)) → le#(x, y) |
| quot#(s(x), s(y)) → quot#(minus(x, y), s(y)) |
| log#(s(s(x))) → log#(s(quot(x, s(s(0))))) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, true, false, quot, log
Strategy
Polynomial Interpretation
- 0: 1
- false: 0
- if_minus(x,y,z): y
- le(x,y): 2y
- log(x): 0
- minus(x,y): x
- quot(x,y): 0
- quot#(x,y): x
- s(x): 2x + 1
- true: 3
Improved Usable rules
| minus(0, y) | → | 0 | | minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |
| if_minus(true, s(x), y) | → | 0 | | if_minus(false, s(x), y) | → | s(minus(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| if_minus#(false, s(x), y) | → | minus#(x, y) | | minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, true, false, quot, log
Strategy
Projection
The following projection was used:
- π (if_minus#): 2
- π (minus#): 1
Thus, the following dependency pairs are removed:
| if_minus#(false, s(x), y) | → | minus#(x, y) |
Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, minus, 0, s, le, false, true, log, quot
Strategy
There are no SCCs!
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, true, false, quot, log
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(0, y) | → | 0 |
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | | if_minus(true, s(x), y) | → | 0 |
| if_minus(false, s(x), y) | → | s(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, true, false, quot, log
Strategy
Polynomial Interpretation
- 0: 0
- false: 3
- if_minus(x,y,z): y
- le(x,y): 0
- log(x): 0
- log#(x): x + 1
- minus(x,y): x
- quot(x,y): x
- s(x): x + 1
- true: 2
Improved Usable rules
| quot(0, s(y)) | → | 0 | | minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |
| minus(0, y) | → | 0 | | if_minus(true, s(x), y) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | if_minus(false, s(x), y) | → | s(minus(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |